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TMUA 2016 · Paper 1 · Question 9 of 20

TMUA 2016 Paper 1 Question 9

Coordinate geometry — Circle from a diameter · translation, reflection, enlargement. Try it first; the answer and a full worked solution are below.

TMUA 2016 · Paper 1Coordinate geometryCircle from a diameter · translation, reflection, enlargement6 options
The line segment joining the points (3,  3) and (7,  5) is a diameter of a circle.

This circle is translated by 3 units in the negative x-direction, then reflected in the x-axis, and then enlarged by a scale factor of 4 about the centre of the resulting circle.

The equation of the final circle is

  1. A(x−2)2 + (y−4)2 = 320
  2. B(x−2)2 + (y+4)2 = 320
  3. C(x−2)2 + (y−4)2 = 80
  4. D(x−2)2 + (y+4)2 = 80
  5. E(x−2)2 + (y−4)2 = 20
  6. F(x−2)2 + (y+4)2 = 20
Show the answer and worked solution
answer · D
  1. A(x−2)2 + (y−4)2 = 320
  2. B(x−2)2 + (y+4)2 = 320
  3. C(x−2)2 + (y−4)2 = 80
  4. D(x−2)2 + (y+4)2 = 80
  5. E(x−2)2 + (y−4)2 = 20
  6. F(x−2)2 + (y+4)2 = 20
The centre is the midpoint (5,  4) and the radius is half the diameter: the diameter has length √42 + 22 = √20, so r = √202 and r2 = 5. Translating 3 to the left moves the centre to (2,  4); reflecting in the x-axis moves it to (2,  −4) and leaves the radius alone. The enlargement is about the circle's own centre, so the centre does not move and only the radius scales, by 4: r2 becomes 16 × 5 = 80. The equation is (x−2)2 + (y+4)2 = 80; scaling r2 by 4 instead of 16 gives the tempting wrong answer 20.